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I am trying to understand the answer to this question.

It was pointed out that this doesn't necessarily hold for non-diagonalizable matrices, so the top answer starts by assuming $A$ and $B$ are diagonalizable, complex matrices. I am comfortable with the first part of this answer:

...we decompose $\mathbb C^n$ as a direct sum of eigenspaces of $A$, say $\mathbb C^n = E_{\lambda_1} \oplus \cdots \oplus E_{\lambda_m}$, where $\lambda_1,\ldots, \lambda_m$ are the eigenvalues of $A$, and $E_{\lambda_i}$ is the eigenspace for $\lambda_i$.

Then, note that each $E_{\lambda_i}$ is invariant under $B$:

If $A v = \lambda_i v, $ then $A (B v) = (AB)v = (BA)v = B(Av) = B(\lambda_i v) = \lambda_i Bv.$

And now, the part that confuses me:

Now we consider $B$ restricted to each $E_{\lambda_i}$ separately, and decompose each $E_{\lambda_i}$ into a sum of eigenspaces for $B$.

How do we know the restriction of $B$ to each $E_{\lambda_i}$ is diagonalizable? I have no guarantee that the eigenbasis of $B$ will be inside $E_{\lambda_i}$, so in general I would expect $B$ to not be diagonalizable. I'm probably missing something here, but this is very non-obvious to me.

Assuming this is true, we can just piece together the resulting eigenspaces to get a common basis of eigenvectors:

Putting all these decompositions together, we get a decomposition of $\mathbb C^n$ into a direct sum of spaces, each of which is a simultaneous eigenspace for $A$ and $B$.

I'm comfortable with this. He then goes on to say

NB: I am cheating here, in that $A$ and $B$ may not be diagonalizable (and then the statement of your question is not literally true), but in this case, if you replace "eigenspace" by "generalized eigenspace", the above argument goes through just as well.

Questions:

  1. Is it true generally that a restriction of a diagonalizable operator to an invariant subspace is also diagonalizable?

  2. His assertion at the end that "if you replace 'eigenspace' by `generalized eigenspace,' the above argument goes through just as well." seems to be outlining a proof to some unknown statement. I'm guessing it is "Matrices commute iff they share a common basis of generalized eigenvectors". Is this correct?

  3. If the answer to 2 is "yes," is the following proof outline correct?

    If $A$ and $B$ commute, so do $(A - \lambda I)$ and $B$. Using the the same manipulation as before we see that the generalized eigenspaces of $A$ are invariant under $B$. Then, the restriction of $B$ to each generalized eigenspace has a Jordan decomposition, and we can piece together these generalized eigenspaces to get a common basis of generalized eigenvectors.

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Noah Caplinger
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    The word basis in your title is misleading. The most you can show is that they have at least one common eigenvector. – Oliver Jones Jan 18 '20 at 23:23
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    As you've pointed out, the proof you've given has holes. Here's a link to a better one: Proof. – Oliver Jones Jan 18 '20 at 23:25
  • @OliverJones, I'm aware the title is misleading, but it is the title of the referenced question. I am asking for clarification on the top answer, which weakens the statement slightly (and has a score of 57, so I'm assuming it is accurate). Could you expound on how the given proof "has holes"? – Noah Caplinger Jan 19 '20 at 02:07
  • You said it yourself; what does it mean to decompose $E_{\lambda_i}$ into a sum of eigenspaces for $B$? Also, there's a more general result where we don't require $A$ and $B$ to be diagonalizable. – Oliver Jones Jan 19 '20 at 02:31
  • If $B$ is diagonalizable on the restriction (which I'm not sure about), then $E_{\lambda_i}$ decomposes as a direct sum of its eigenspaces (that's the whole point of diagonalization). My question is about how you conclude $B$ is diagonalizable on the restriction. – Noah Caplinger Jan 19 '20 at 02:37
  • If you check the original question you'll see that there's no mention of the matrices being diagonalizable. – Oliver Jones Jan 19 '20 at 09:37
  • In the final paragraph of the top answer: "NB: I am cheating here, in that A and B may not be diagonalizable." If you are referring to the fact that the answer doesn't mention that the restriction of $B$ is not diagonal, he does mention that it decomposes as a direct sum of eigenspaces, which is equivalent to diagonalizability. In any case, the question has 32k views, with the answer 57 upvotes. I have reason not do doubt this answer. – Noah Caplinger Jan 19 '20 at 15:52
  • This is a well-known result in linear algebra and I've given you a link to one proof. There is also a nice proof in Horn and Johnson's book "Matrix Analysis". – Oliver Jones Jan 19 '20 at 23:28
  • Sorry if I'm misunderstanding, but the proof you've linked to is about having a single common eigenvector. Is the top answer to the other question wrong? If it is, could you explain in detail why? If not, could you explain the step I was having trouble with? – Noah Caplinger Jan 20 '20 at 00:15
  • The original question is worded badly and so it's unclear what's being asked. The hole in the first proof can be fixed by the fact that a diagonalizable matrix restricted to an invariant subspace is also diagonalizable. Here's a link to a proof: https://math.stackexchange.com/a/62340 – Oliver Jones Jan 20 '20 at 08:48

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Answer to this question:

  1. Is it true generally that a restriction of a diagonalizable operator to an invariant subspace is also diagonalizable?

If $B$ is diagonalized by the matrix $P$ then the restriction of $P$ to $E_{\lambda_i}$ diagnolized the restriction of $B$ by

  1. I'm guessing it is "Matrices commute iff they share a common basis of generalized eigenvectors". Is this correct?

Yes, if A,B commute, then one can find a basis that is both genearlized eigenvectors of A and B. Your proof looks good to me.

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